Let’s say that a set *contains an singular matrix* if there exists an matrix with determinant zero whose entries are distinct elements of . For example, the set of prime numbers does not contain any singular matrices; however, for every it contains infinitely many singular matrices.

I don’t know of an elementary proof of the latter fact. By a 1939 theorem of van der Corput, the set of primes contains infinitely many progressions of length 3. (Much more recently, Green and Tao replaced 3 with an arbitrary .) If every row of a matrix begins with a 3-term arithmetic progression, the matrix is singular.

In the opposite direction, one may want to see an example of an infinite set that contains **no** singular matrices of any size. Here is one:

Continue reading